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A Mind Boggling Probability Problem

Q. A man comes up to you on the street and says: I have two children. At least one of them is a boy. What is the probability that the other child is also a boy?

This question is much trickier than it sounds!

Try to solve it yourself before reading on.

P.S. definitely read till the fourth part of this problem, "The Unlikely Boy". That is the mind-boggling bit.

Anyone can solve this question, you don't need special math or programming knowledge.

Hint: the answer is not 50%.

The wrong answer

Having a boy shouldn't affect the odds of having a boy or a girl. The odds are still 50/50. So the answer is 50%.

This part is right:

Having a boy shouldn't affect the odds of having a boy or a girl. The odds are still 50/50.

This part is wrong:

So the answer is 50%.

The right answer

Since there are two children, here are all the possibilities of what they could be (G = girl, B = boy). The first letter is for the first child, the second letter is for second child.

So he could have two girls, older girl and younger boy, younger girl and older boy etc.

Since we know at least one is a boy, he can't have two girls:

Out of the three cases, only one of them is BB (two boys). So the odds that the other one is also a boy, are 1 in 3.

That makes sense, but the wrong reasoning also made sense. I'm confused.

Suppose the man in the street asked:

I have two kids, what are the odds I have 2 boys?

Then you would naturally list out all four cases:

And reply 1 in 4.

But the wording of this problem tricked you into just considering 2 cases:

The wrong solution only takes the BB and BG cases into account, and not the GB case.

Since there is no way to differentiate between the two kids, you have to calculate the possibilities for both kids together:

Let's see the next problem.

The Older Boy

Q. A man comes up to you on the street and says: I have two children. The older one is a boy. What is the probability that the other child is also a boy?

Answer

Lets list out the four possibilities again. In each case, the older child is listed first:

Since the older one is a boy, we can eliminate the case where the older one is a girl:

One out of two cases is BB, so the answer is 1/2!

Another explanation:
This time there is a way to differentiate between the two kids. So this logic is now okay:

There's a 50% chance the other child is a boy.

The Boy Who Was Standing There

Q. A man comes up to you on the street and says: I have two children. One is the boy standing here next to me. What is the probability that the other child is also a boy?

Answer

Let's list out the possibilities again. We started with four:

But which child is standing in front of you? You have two options:

If it is boy-boy, it could be the older boy or the younger boy standing in front of you.

To make it less cluttered I will replace the text with a yellow highlight:

The yellow highlight indicates which child is standing in front of you.

Similarly for all of these you have two options:

So really there are eight possibilities, not four:

Since he told us that the child standing next to him is a boy, we can eliminate some possibilities:

Now 1/2 of the options are BB, so the odds are 1/2 that the other child is also a boy.

The Unlikely Boy

This lady was just out for a nice probability-free walk

This is the crazy part.

Q. A man comes up to you on the street and says: I have two children. One of them is a boy who was born in the summer. What is the probability that the other child is also a boy?

Hints:

The answer is not 1/2.

"who was born in the summer" gives you extra information.

Suppose the man had said "one of them is a boy who was born on Diwali". Would that make it more or less likely that he has two boys?

Just like in problem 1, we have no way of differentiating between the two boys. It is possible they are both born in the summer!

Answer

This one is hard! Once again we have four possible cases, and we can eliminate GG:

Now we also need to take the season into account. Here's what that would look like:

So with two kids, there are 16 possible combinations: you could have child 1 born in the spring and child 2 born in the fall, or child 1 born in the winter and child 2 born in the spring etc etc.

Now let's list out all the cases for GG, GB, BG and BB, taking the seasons into account:

(again there are no options for GG because we need at least one boy, since the man told us one child is a boy).

That extra information of "born in the summer" generated a lot more cases! We now have 16 * 3 = 48 cases to consider. But we only need to consider the cases where there is a boy born in the summer:

There are 15 cases where a boy is born in the summer:

Out of those, there are 7 cases involving two boys:

So the answer is 7/15.

The Diwali Boy

In the hints I said:

Suppose the man had said "one of them is a boy who was born on Diwali". Would that make it more or less likely that he has two boys?

The odds of that happening are so small! But if you have two boys, you have a much better chance of this low-probability event happening, because you got two shots at it. If you had 100 boys, you would have an even better chance of having a boy who was born on Diwali.

Even though it seemed useless, that extra bit "born in the summer" actually gave you some useful information. Since we know this low-probability thing occured, so we can work backwards and say, there's a higher chance you have two boys, because someone with two boys is more likely to have a boy born in the summer. So the odds increase from 1/3 to 7/15.

Further Reading

I got this problem from Probability: For the Enthusiastic Beginner by David Morin. It is super easy to read and has lots of diagrams and examples (I also recommended it last year). He adds more detail on this problem and covers the other classic problems in probability. I highly recommend it!